Background:
Proposition 1: Let $L$ be a group and $G$ and $H$ be two subgroups. Then, the following statements are equivalent:
$L,G$ and $H$ satisfy the following relations: $L=GH, G\cap H=\{1\},$ and $|G,H|=\{1\}.$
$L$ is the direct product of $G$ and $H,$ i.e. $L\cong G\times H.$
Proof: Write
$$G=\{(g,1),g\in G\}$$
and
$$H=\{(1,h),h\in H\}$$
Then, obviously $L=GH$ and $G\cap H=\{1\}.$ For $[G,H]=\{1\},$ we have
$$(g,1)(1,h)(g,1)^{-1}(1,h)^{-1}=(1,1).$$
Questions:
I just have a quick question about: $(g,1)(1,h)(g,1)^{-1}(1,h)^{-1}=(1,1).$, specifically in $(g,1)^{-1}(1,h)^{-1}$, does $(g,1)^{-1}=(1,g^{-1})$ and similarly $(1,h)^{-1}=(h^{-1}, 1)$?
Thank you in advance
Let $(g,h)\in G\times H$. Consider $(g^{-1}, h^{-1})$. We have
$$\begin{align} (g,h)(g^{-1}, h^{-1})&=(gg^{-1}, hh^{-1})\\ &=(e_G, e_H)=e_{G\times H}\\ &=(g^{-1}g,h^{-1}h)\\ &=(g^{-1}, h^{-1})(g,h).\\ \end{align}$$
Therefore, by uniqueness of inverses, we have $(g,h)^{-1}=(g^{-1}, h^{-1}).$